Rotating frames desynchronization

[JVNY]

I agree that this is essentially synchronizing the clocks to the axis, which is not the goal. As you object earlier,

. . . Synchronizing the clocks on the equator according to a master clock at the center . . . just makes the clocks agree on simultaneity of events relative to the master clock, not on simultaneity of events relative to their own rest frames . . .

The goal is to determine whether rim observers can agree on the simultaneity of events relative to their rest frame (the rim), not relative to some other location (like the axis). We have concluded that observers at rest on the inertial platform can agree on the simultaneity of events relative to their platform without using synchronized watches, and without even using coherent watches. Can rim observers do the same? If not, why not?

 

[WannabeNewton]

Sorry I should have been more careful when I said what I did in that quote. What I meant was simultaneity amongst the clocks on the rim of the rotating disk as defined using the global time coordinate of the central clock's inertial frame will not give us Einstein simultaneity amongst the clocks on the rim. But certainly the clocks on the rim will agree on simultaneity of events on the rim if simultaneity is defined using light signals from the central clock.

 

[PAllen]

I agree that this is essentially synchronizing the clocks to the axis, which is not the goal. As you object earlier,



The goal is to determine whether rim observers can agree on the simultaneity of events relative to their rest frame (the rim), not relative to some other location (like the axis). We have concluded that observers at rest on the inertial platform can agree on the simultaneity of events relative to their platform without using synchronized watches, and without even using coherent watches. Can rim observers do the same? If not, why not?

Let me first ask your preference: Do you want to consider a ring or a disk? What I mean is, suppose rim observers want to do something analysis to simultaneity via lightning bolt strikes. Do the see the light moving directly via vacuum (ring), or do we pretend light is forced to follow effectively fiber optic paths around the rim (which is what would be needed for seeing around a disc)?

 

[WannabeNewton]

Sorry I should have been more careful when I said what I did in that quote. What I meant was simultaneity amongst the clocks on the rim of the rotating disk as defined using the global time coordinate of the central clock's inertial frame will not give us Einstein simultaneity amongst the clocks on the rim. But certainly the clocks on the rim will agree on simultaneity of events on the rim if simultaneity is defined using light signals from the central clock.

Thankfully, I found a rigorous mathematical treatment of this.

See section IV of the following paper: http://arxiv.org/pdf/gr-qc/0405139v2.pdf as well as the paragraph directly above section 3 (conclusion) of the following paper: http://arxiv.org/pdf/gr-qc/0506127.pdf

 

[pervect]

I would prefer a concrete example to clarify this conceptual mess in my head. So clocks on Earth that are at rest, when we consider them as the frames of reference, disagree on simultaneity. I get this, but what is the criteria for this. All clocks on the line of rotation have different perspectives on simultaneity? How does their perspective differ? There are many questions in my head and I doubt maths would help it since I'm not an excellent mathematician like most of you guys. Can you give me an example that is based with some clocks on earth, or something like that?

I'm not quite sure I understand what your confusion is, so it's hard to address.

Let me add a few things.

A clock that is "at rest" as in having constant lattitude and longitude is not at rest relative to a hypothetical non-rotating inertial frame based at the center of the Earth.

Einstein synchronization is based on a non-rotating and inertial (or nearly inertial) frame of reference. You were asking about "criterion" - the "criterion" for Einstein clock synchronization is is that clocks moving at a different velocity in an inertial (or nearly so) frame of reference have different concepts of simultaneity.

THe clocks "at rest" (as in having constant lattitude) all have different velocities, and hence all have different notions of "Einstein simultaneity" for points near them. Is this what you were asking about? I wasn't quite sure.

You are also asking about how we actually keep time on Earth. YOu can read about "atomic time", aka TAI time on the wikipedia. You'll note that it does NOT use Einstein synchronization, which is logical because TAI covers the whole Earth without any discontinuities, and as it's been remarked this isn't possible with Einstein clock synchronization.

All clocks on the geoid all run at the same rate, and the TAI standard defines a notion of simultaneity that is not the same as Einstein's.

THis has some consequences to the laws of mechanics, etc - I'm not going to go into detail unless this is one of the things you're interested in. Well, I will say one thing. The Einstein notion of simultaneity (which is the one we are NOT using on the Earth) is really the best/simplest one mechanics, so expect a few surprises when you write the laws of mechanics on the Earth using TAI time. Going into the details would probalby just be confusing until we get the rest of your confusions sorted, I think.

You were also asking about how we order events. This doesn't have anything to do with how we define simultaneity in relativity. Ordering of events in an observer independent manner is always done with light cones, because ordering events according to their coordinates depend on what coordinates you use (I think this is obvious?).

Briefly, I can say that "events in the past light cone are in the past" and "events in the future light cone are in the future" isn't clear enough, I suppose I could go on in more detail. But since I don't know what you're confused about and what you're interested in, it would seem to be better to wait for a question.

So you asked about a bunch of different things, and I tried to answer each "tangent". I hope that clarifies things rather than confuses you more, but I can't quite figure out what you want to know,.

 

[JVNY]

Let me first ask your preference: Do you want to consider a ring or a disk? What I mean is, suppose rim observers want to do something analysis to simultaneity via lightning bolt strikes. Do the see the light moving directly via vacuum (ring), or do we pretend light is forced to follow effectively fiber optic paths around the rim (which is what would be needed for seeing around a disc)?

I would prefer to consider light traveling around the rim (for example by traveling along a mirrored interior surface of the rim, or through fiber optic cables laid along the rim, or the like) if that is instructive. Thanks.

 

[WannabeNewton]

I would prefer to consider light traveling around the rim (for example by traveling along a mirrored interior surface of the rim, or through fiber optic cables laid along the rim, or the like) if that is instructive. Thanks.

The simplest simultaneity convention that makes use of light signals along the rim to and fro observers is Einstein simultaneity and as you know this will not give rise to a valid global time coordinate for the family of observers on the rim. On the other hand if you use the Einstein time of the inertial frame fixed to the symmetry axis as the global time coordinate for the family of observers on the rim then you will get a consistent global simultaneity convention and it will just be given by the simultaneity surfaces of the observer at the center of the disk. The observers on the disk will agree on simultaneity of events anywhere and everywhere as per this convention-it's trivially transitive because it's just the synchronous time of an inertial frame. Keep in mind this simultaneity convention only works because of axial symmetry.

 

[FactChecker]

If you had lots of clocks all around the Equator, at rest relative to the Earth's surface, and used Einstein synchronisation to sync the 2nd clock to the 1st clock, then the 3rd to the 2nd, then the 4th to the 3rd, and so on all round the Equator until you got back where you started, you would find that the last clock and the 1st clock, which are side-by-side, would be out of sync by about 200 nanoseconds.\frac{ \left( \frac{40 \times 10^6}{24 \times 60 \times 60} \right) \times \left( 40 \times 10^6 \right) } {\left( 3 \times 10^8 \right) ^2} \approx 2 \times 10^{-7}For an object rotating much faster than the Earth, the effect would be greater.

I can understand the clocks being out of sync with a clock at the center of the earth, but why out of sync with other clocks at the same distance from the axis. Aren't they all slowed an identical amount versus the clock at the center? Is this an effect related to what causes precession of Mercury's orbit?

 

[WannabeNewton]

Aren't they all slowed an identical amount versus the clock at the center?

That's not enough to guarantee global Einstein synchronization of the clocks. Just because they are slowed by the same amount doesn't mean Einstein synchronization will be transitive for the clocks on the periphery of the disk. The point is that in a rotating frame Einstein synchronization between clocks at rest in the frame will never be transitive and will therefore lead to time discontinuities.

This is due to a well known result in relativity which states that given a family of standard clocks each following an integral curve of a time-like vector field $\xi^{\mu}$, a necessary condition for the standard clocks to be Einstein synchronizable is that $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} = 0$. If you've ever seen Frobenius' theorem in differential topology then it will be evident why this is so.

However for clocks at rest in a rotating frame we have $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} \neq 0$ and so Einstein synchronization is impossible for these clocks.

This is why we compensate by synchronizing the disk clocks to the time of the central clock instead. They will no longer be standard clocks, rather they will be coordinate clocks in the rest frame of the rotating disk, but they will at least be synchronized and share common global simultaneity surfaces. This works of course because the central clock follows an integral curve of the time-like vector field $\nabla^{\mu}t$ where $t$ is the time coordinate in the rest frame of the rotating disk and hence the time read by the central clock and the inertial clocks at rest with respect to it-we see trivially that $\nabla_{[\gamma}t\nabla_{\mu}\nabla_{\nu]}t = 0$.

EDIT: See chapter 3 of Gron's notes: http://www.uio.no/studier/emner/matnat/fys/FYS4160/v06/undervisningsmateriale/kompendium.pdf

 

[pervect]

I can understand the clocks being out of sync with a clock at the center of the earth, but why out of sync with other clocks at the same distance from the axis. Aren't they all slowed an identical amount versus the clock at the center? Is this an effect related to what causes precession of Mercury's orbit?

The precession of Mercury's orbit is a GR effect - the clock synchronization issue appears within special relativity. So I don't think they are directly related.

It confuses a lot of people who hold onto the idea of "absolute time".

A close study of "Einstein's train", the issue where two lightning bolts strike the front and rear of the train simultaneously in one frame, and not simultaneously in the other, is the key to understanding the effect.

Einstein's original argument can be found at http://www.bartleby.com/173/9.html - but it doesn't give any numbers. You can get the numbers directly from the Lorentz transform though:

$t' = \gamma ( t - \beta x/c)$

where $\beta = v/c$ and $\gamma = 1 / \sqrt{1-\beta^2}$

We can break the above equation for t' into two parts or terms

term 1: $\gamma t$
term 2: $-\beta \gamma x/c$

term 1 represents "time dilation:
term 2 represents "the relativity of simultaneity" and is responsible for the issue under discussion

Given the numerical approach above, you basically apply the result from the Einstein train thought experiment around a closed loop of "trains", summing together the second term around the loop.

 

[FactChecker]

term 2 represents "the relativity of simultaneity" and is responsible for the issue under discussion

Given the numerical approach above, you basically apply the result from the Einstein train thought experiment around a closed loop of "trains", summing together the second term around the loop.

Ha! I like that.

 

[JVNY]

The simplest simultaneity convention that makes use of light signals along the rim to and fro observers is Einstein simultaneity and as you know this will not give rise to a valid global time coordinate for the family of observers on the rim. On the other hand if you use the Einstein time of the inertial frame fixed to the symmetry axis as the global time coordinate for the family of observers on the rim then you will get a consistent global simultaneity convention and it will just be given by the simultaneity surfaces of the observer at the center of the disk. The observers on the disk will agree on simultaneity of events anywhere and everywhere as per this convention-it's trivially transitive because it's just the synchronous time of an inertial frame. Keep in mind this simultaneity convention only works because of axial symmetry.

Per our follow up discussion on another thread ("Synchronizing rotating clocks"), there is another convention that does work by sending signals only around the rim, see post 70 linked here: https://www.physicsforums.com/showthread.php?t=732892&page=4. Interestingly, even though the signals only go around the rim, they likely also synchronize the clocks on the rim to an axis centered inertial coordinate system -- despite there being no signals sent from the axis or from any other point at rest with respect to the axis.